## Probabilities of "favorite-betrayal lesser-evil" (FBLE) situations in 3-candidate ranked-ballot elections

Probabilistic model: All elections equally likely. That is, with V voters, there are 6V possible elections since each voter can vote in 6 ways: A>B>C, A>C>B, B>A>C, B>C>A, C>A>B, C>B>A. (We disallow "truncated ballots" and "ranking equalities.") We shall consider the large-V limit V→∞.

Definition of "FBLE situation": Call the election-winner A. An FBLE situation then occurs when some C>B>A voters, by switching to B>C>A ("betraying their favorite" C) can make the "lesser evil" B win (an outcome they prefer). Note: We shall abbreviate C>B>A votes by "CBA," et cetera, in what follows.

### Condorcet

Condorcet's FBLE probability is 8.78% (but arguably 100%). Proof sketch: An FBLE situation occurs only in the event there is a Condorcet cycle A>C>B>A with A suffering the weakest defeat (by C); the CBA voters by switching to BCA cause B to defeat C and become a Condorcet winner (or at least to only be beaten maximally-weakly by C). (Concrete example but with letters changed.)

Rather non-obviously, a Condorcet cycle happens with asymptotic probability (3*arccos(1/3)-π)/(2π)≈8.78%. (The obvious guess, 25%, is wrong, see puzzle #7 for an explanation of why the truth is 8.78%.) Due to symmetry, the sum of all the pairwise votes behaves like a 3-dimensional normal distribution, but not a spherically-symmetric one due to correlations. Given such a cycle, FBLE happens (i.e. there are enough CBA voters to cause C to become a Condorcet winner after they switch) with asymptotically 100% probability.

Strengthening: If could, however, be argued that if the CBA voters are convinced C cannot win, then their betrayal by switching to BCA cannot hurt them but with 8.78% probability will help them. I.e, if convinced C cannot win, then this betrayal is a strategically sometimes-wise and never-unwise move with 100% probability.

Discussion: For the voters considering betraying the third-party candidate C (who is their honest favorite) to make the lesser-evil B win, the question is this. Which is more likely?

1. By this betrayal I will cause C to lose whereas otherwise he would have won. (Bad! Backfire!)
2. By this betrayal I will cause B to win whereas otherwise the greater-evil A would have won. (Good! Success!)
If the probability that third-party candidates can win is well below 8.78% (which is certainly true in present-day USA politics), then the voter is fully justified, 100% of the time (!), in deciding to betray. If voters act that way, then third party candidates will never win – justifying their decision! – and then self-reinforcing 2-party domination is guaranteed, just like under the plurality system that Condorcet was supposed to "improve over."

Mind you, it is not clear Condorcet voting really will lead to self-reinforcing 2-party domination. That experiment has never been done and it might be that "8.78%" is small enough that this trap can be escaped. We do not know. But certainly the risk is there. And since every IRV country is 2-party dominated and for IRV the magic number is 19.6% rather than 8.78%, we know 19.6% is enough to do the job.

Another question: 8.78% is the FBLE and cycle probability in our mathematical model, but what is that number in real life? The answer is unknown, but based on surveying the few "real life" Condorcet elections that have been conducted so far, I am willing to say that it is between 0 and 20%, but unwilling to be any more precise than that.

### Instant Runoff Voting

Instant Runoff Voting's FBLE probability is 19.6% (but arguably 100%). Proof sketch: An FBLE situation occurs only in the event B is eliminated in the first round and then A beats C in the second and final round to take the prize. If enough CBA voters switch to BCA then C is eliminated in the first round instead (with asymptotic 100% probability there are enough CBA voters to cause this), and then hopefully B beats A. (Concrete example; Another type; Another.) The 100%s are coming from the fact that, with asymptotically 100% probability, there are about V/6 voters of each kind, while the differences between any two candidates' vote counts are only of order √V, which is far smaller.) The "hope" that B will beat A is satisfied with probability P, where P is the probability that given independent standard normal (Gaussian) random variables S,T,W,X,Y,Z, (the original vote counts for ABC, ACB, CAB, CBA, BAC, and BCA respectively) and given that S+T>Y+Z and W+X>Y+Z, we have that S+T+W<Y+Z+X. (The Gaussians are justified by the central limit theorem.)

Let's do that again in slo-mo: (1) S+T>Y+Z and (2) W+X>Y+Z make sure B is eliminated in the first round. Now our design goal is after the CBA→BCA switch, to make C lose and B win. After Q voters switch where necessarily (3) 0<Q≤X then we want (4) S+T+Q>W+X to make A beat B in the first round, and we want (5) Y+Z+2Q>W+X to make B beat C in the first round. Finally we want (6) S+T+W<Y+Z+X to make B beat A in the second round. Here the reader notes that we mentioned (1) and (2) and (6) above but did not mention (3) and (4) and (5). Huh? That is because we may as well take Q=X since that only makes everything we said (if it is true) more true. In that case (4) becomes just S+T>W, while (5) becomes just Y+Z+X>W. These two demands can be dropped because in our random model they both are satisfied with 100% probability in the V→large limit. So we don't have to mention these two.

More simply but equivalently, P is the probability that given independent standard normal (Gaussian) random variables X,Y,Z, we have that |X|<Y/√2, whereupon by considering polar coordinates instead of XY, we see that P=arctan(1/√2)/π=0.195913276...

That in slo-mo: The point is that in the original 6-variable formulation, S and T only occur in the combination S+T, which we may regard as a different normal random variable with √2 times larger standard deviation. Similarly for Y+Z. We may also treat W+X and W-X as independent normal random variables with enlarged standard deviation. The problem then becomes to find the probability that "given F>G and H>G that F<G+L" where F,G,H,L are i.i.d. standard normal random variables. But then write |X|=F-G and Y=L√2 to get the final formulation in only 3 variables.
A confidence builder: Just in case all those manipulations zoomed by a little too quickly for those not used to working with Gaussian random variables... I have also confirmed this to 4 decimal places just by Monte Carlo integration for the original 6-dimensional scenario. The fact both the Monte Carlo and exact calculations agree, is convincing evidence the reduction from a 6-variable to 3-variable problem, was valid.
Finally, we have not explained in detail the case analysis showing why this scenario is the only way to get FBLE – but if you don't buy that, then this still shows 19.6% is a lower bound on the FBLE probability.

Remark: IRV also allows a (more bizarre) form of favorite betrayal, which has not been counted in the above "19.6%" because it works differently: A wins, then the C>B>A voters betray C in order to make C win! Here's an example of that. (If these were also to be counted, then presumably "19.6%" would increase?)

There also is an alternative "Dirichlet" probabilistic model in which you get 20.2% not 19.6%.

Strengthening to "100%": It could, however, be argued that even if the "hope" is not satisfied (i.e. the betrayal fails to elect B; A still wins) our CBA voters still were not hurt by making this switch. Therefore, if they were convinced C could not win, then betraying C would be a strategically sometimes-wise and never-unwise move with 100% probability.

Indeed, there is no need for the CBA voters to be convinced C cannot win; it suffices for them merely to believe C's winning chances are well below 19.6% so that their move is far more likely to pay off than being honest.

That belief is justified. In US senate, house, and presidential elections, the winning percentage for third-party candidates is well below 1% historically. In Australia's House, elected via IRV, there currently are zero third-party-held seats. (In both the US and in Australia's House it appears being "independent" is better for your chances than being in a third party; and being a former major-party member helps a lot if so.)

Alternative (more realistic?) mathematical model #1: We can change our randomized assumption (which we had chosen for maximum simplicity rather than maximum realism) from "all elections equally likely" to, e.g, "all votes equally likely except that the probabilities that A,B,C will be middle-ranked of the three in that vote are 30%, 30%, and 40% respectively" where C is the 3rd-party candidate. Then in IRV as #voters→∞, C's probability of winning is provably exponentially tiny so that Joe Voter is justified in assuming C only a very tiny chance t of winning. Indeed C only has a tiny chance of merely surviving the first round. However, Joe reasons, if Joe and friends by honestly-ranking C top do manage to make C survive the first round, then that will almost certainly happen only at the cost of eliminating Joe's second-favorite candidate A. If the A votes then transfer equally to C and B (which in "1-dimensional politics" with C A B arranged along a "line" in that order, seems likely) then C will almost certainly still lose, and will have deprived A of victory in the process.

The idea then would be that the behavior of mid-ranking the 3rd party candidate would be self-reinforcing in IRV: an assumption of a slight bias that way like we just made (40% versus 30%, or situation like this example), then leads to it being strategically wise for Joe Voter to do it, leading to a larger bias that way, etc. – positive feedback, self-reinforcing 2-party domination.

Alternative (more realistic) mathematical model #2: We can change our random-model to be "random selection from a set of situations resembling this example."

This is somewhat more "realistic 1-dimensional politics," and also somewhat resembles a situation in which Joe Voter has seen some pre-election polling, and it leads to betraying C being the right strategic move.

### Summary:

This discussion presumably is the underlying theoretical explanation for the fact that all three IRV countries (Ireland, Australia, Malta, and with the recent addition of Fiji) historically have been 2-party dominated.

Evidently the IRV voting method leads to 2-party domination, just like the flawed plurality system that method was supposed to "fix." It also is plausible that Condorcet methods with full rankings as ballots also will lead to 2-party domination.

So anybody who is interested in third parties ever having a chance, would be advised not to foolishly advocate IRV, but instead would be advised to advocate Range Voting (which experimentally favors all third parties far more than either plurality or approval, incidentally, see the CRV web site).

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